Joined: 4/11/2006
Posts: 487
Location: North of Russia :[
Sorry about the resubmission guys, I wish I had caught Nitrodon sooner who actually figured out exactly how the Dark King cure glitch works!
That's what forum topics at "Games" subforum are for =) To post WIPs and get help ^^
Sorry, as I have never played I won't be able to understand if this movie good or bad, So I won't be even watching it, but still...
LAST BOSS GLITCH SOUNDS TO BE COOL :D (read about it on IRC when you discussed it with Nitrodon).
I noticed almost all of your fights required both player 1 and 2 to attack using their standard weapons instead of using a magic attack on all enemies at once. Was using a single magic attack (such as white) actually slower in most cases?
Nobody reads submission texts... And it says: "I really didn't start saving time until I got the Knights Sword. ... Second it made me more powerful, so I could swing my sword which takes significantly less time then any magic attack (except Exit)."
Joined: 4/11/2006
Posts: 487
Location: North of Russia :[
I think this game TAS can be impruved on and on, with different stragies, pathes, glitches etc, but it takes very many time to record movie to gain just few seconds, so... I am wondering, if it's worth it? I still like Mew run much more ^^
Joined: 4/11/2006
Posts: 487
Location: North of Russia :[
I would like to see leveling up using oddish 230 battle at least once... Rare candies can be used afterwards =)
And primo, when movie is finished, can you make "bonus version" with catching as many different glitch pokemon as possible after getting diploma? Not for submission may be, but it would be really nice extra...
Joined: 4/11/2006
Posts: 487
Location: North of Russia :[
Only 136~ I am below this forum's average? T_T But anyways at least two questions did not display correctly with disabled graphics even when I tried to make it show those pictures ^^
Joined: 4/11/2006
Posts: 487
Location: North of Russia :[
I have killed few monsters in Dawn of Sorrow on NDeSmuMe... I did not see them though, only health bar was displayed on bottom screen. ingame menu, stats and maps worked nice, start menu did not display at all (
Joined: 4/11/2006
Posts: 487
Location: North of Russia :[
I will check it tomorrow =) The only thing I can say now is that at "psycho" speed ball can move VERY fast, ecpecially if it gets between wall and closing flipper in upper part of board... tilting is the thing to play around with~ I guess I can get 6x bonus in no time with it D
Joined: 4/11/2006
Posts: 487
Location: North of Russia :[
I have 2 questions
1) Do you think it can make entertaining TAS?
2) What should be its objective?
I can see getting Jackpots in all four boards (multi-board mode) with largest multiplier using 1 ball on psycho speed as an objective... And on easy mode, as it reduces number of repeatative actions needed to perform to get different bonuses...
Joined: 4/11/2006
Posts: 487
Location: North of Russia :[
DrJones wrote:
Prove that no number on the sequence:
5
15
115
1115
... (and so on)
can be divided by seven.
It's simple...
N1 = 5
N1 mod 7 = 5
Nn = (Nn-1 - 4) * 10 + 5
Nn = 10*Nn-1 - 35
35 mod 7 = 0
10*Nn-1 can not be divided by 7 as Nn-1 can not and 10 can not
10*Nn-1 mod 7 = (10*Nn-1 - 35) mod 7
therefore for any n Nn can not be divided by 7
Joined: 4/11/2006
Posts: 487
Location: North of Russia :[
Okay
let's prove that k^(1/(k-1)) takes only one integer value when k>1
f(k) = k^((k-1)^-1)
lim f(k) = e
k -> 1
lim f(k) = 1
k -> infinity
f'(k) = -k^(k-2) * (k-1)^-3
k>1 -> k^(k-2) > 0
k>1 -> (k-1)^-3 > 0
f'(k) < 0 when k > 1
therefore it takes values from e downto 1 always decreasing. We can conclude it takes values from range (1;e). 2 is the only integer there.
Joined: 4/11/2006
Posts: 487
Location: North of Russia :[
a^b = b^a (a < b, a and b are integers)
b = a^k (k>1)
a^(a^k) = (a^k)^a (=a^(ak))
ak = a^k
a = k^(1/k-1)
so k^(1/k-1) should be integer with k>1...
but (k^(1/k-1)) when increasing k goes to 1 from above, and the ony integer value it takes is 2 (you can build graph (y(x) = x^(1/x-1)) and see yourself)
a = 2 ^ (1/(2-1)) = 2
b = 2^2 = 4
so, a ^ b = b ^ a only when a = 2 and b = 4 or b = 2 and a = 4
P.S. Thanks, solving this question was quite interesting )
Joined: 4/11/2006
Posts: 487
Location: North of Russia :[
I wish i had my old keyboard with "turbo" button... I could just press turbo+F7 (or turbo+f1? I don't remember which is the fastest and which is slowest) and hold down enter for 100+ results D
